Question

Find the orthogonal projection of $\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$onto the subspace of ${\mathbf{ℝ}}^{\mathbf{4}}{\mathbf{ℝ}}^{\mathbf{4}}$ spanned by $\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]$ and $\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]$.

Short answer

The orthogonal projection is$pro{j}_v\overrightarrow{x}=\left[\frac{3}{4}\frac{1}{4}-\frac{1}{4}\frac{1}{4}\right]$ .

Step-by-step solution

Determine the orthonormal basis.

Consider a set $span\left\{{\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]\right\}$spanned ${\mathrm{ℝ}}^4$and a vector $\overrightarrow{x}=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$.

If the vector localid="1659950736693" $\overrightarrow{\mathbf{x}}$is perpendicular $\overrightarrow{\mathbf{y}}$then $\overrightarrow{\mathbf{x}}\mathbf{.}\overrightarrow{\mathbf{y}}\mathbf{=}\mathbf{0}$.

By the theorem of orthogonal basis, the orthogonal ${\overrightarrow{u}}_1$is defined as follows:

${\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||}$

Substitute the value $\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$for ${\overrightarrow{v}}_1$in the equation ${\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||}$as follows:

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||} \\ {\overrightarrow{u}}_1=\frac{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]||} \\ {\overrightarrow{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\overrightarrow{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{||{\overrightarrow{v}}_1||} \\ {\overrightarrow{u}}_1=\frac{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]||} \\ {\overrightarrow{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\overrightarrow{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

${\overrightarrow{u}}_1=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$

Similarly, the orthogonal${\overrightarrow{u}}_2$ is defined as follows.

${\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||}$

Substitute the value$\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]$ for${\overrightarrow{v}}_2$ in the equation${\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||}$ as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2}{||{\overrightarrow{v}}_2||} \\ {\overrightarrow{u}}_2=\frac{\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]}{||\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right]||} \\ {\overrightarrow{u}}_2=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \\ {\overrightarrow{u}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ -1\\ -1\end{array}\right] \\ {\overrightarrow{u}}_2=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right] \end{gathered}$$

Similarly, the orthogonal${\overrightarrow{u}}_3$ is defined as follows.

${\overrightarrow{u}}_3=\frac{{\overrightarrow{v}}_3}{||{\overrightarrow{v}}_3||}$

Substitute the value$\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right]$ for${\overrightarrow{v}}_3$ in the equation${\overrightarrow{u}}_3=\frac{{\overrightarrow{v}}_3}{||{\overrightarrow{v}}_3||}$ as follows.

$$\begin{gathered} {\overrightarrow{u}}_3=\frac{{\overrightarrow{v}}_3}{||{\overrightarrow{v}}_3||} \\ {\overrightarrow{u}}_3=\frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2+{\left(-1\right)}^2}}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \\ {\overrightarrow{u}}_3=\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\overrightarrow{u}}_3=\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \\ {\overrightarrow{u}}_3=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right] \end{gathered}$$

Therefore, the orthogonal basis is$span\left\{{\overrightarrow{u}}_1=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right],{\overrightarrow{u}}_2=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right],{\overrightarrow{u}}_3=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right]\right\}$ .

Determine the orthogonal projection

By the theorem of orthogonal projection, the orthogonal projection of vector$\overrightarrow{x}$is defined as follows.

$pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2+\left({\overrightarrow{u}}_3.\overrightarrow{x}\right){\overrightarrow{u}}_3$

Substitute the value$\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]$for${\overrightarrow{u}}_1$, $\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]$for${\overrightarrow{u}}_2$, $\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right]$for${\overrightarrow{u}}_3$and$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$for$\overrightarrow{x}$in the equation$pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2+\left({\overrightarrow{u}}_3.\overrightarrow{x}\right){\overrightarrow{u}}_3$as follows.

localid="1659952933885" $$\begin{gathered} pro{j}_v\overrightarrow{x}=\left({\overrightarrow{u}}_1.\overrightarrow{x}\right){\overrightarrow{u}}_1+\left({\overrightarrow{u}}_2.\overrightarrow{x}\right){\overrightarrow{u}}_2+\left({\overrightarrow{u}}_3.\overrightarrow{x}\right){\overrightarrow{u}}_3 \\ pro{j}_v\overrightarrow{x}=\left\{\left(\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]+\left(\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right] \\ \left(\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]\right)\left[\begin{array}{cccc}\frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\end{array}\right]\right\} \\ pro{j}_v\overrightarrow{x}=\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\end{array}\right]+\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]+\left(\frac{1}{2}\right)\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& -\frac{1}{2}& -\frac{1}{2}\end{array}\right] \\ pro{j}_v\overrightarrow{x}=\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} pro{j}_v\overrightarrow{x}=\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& \frac{1}{4}& \frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& \frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}\end{array}\right]+\left[\begin{array}{cccc}\frac{1}{4}& -\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \\ pro{j}_v\overrightarrow{x}=\left[\begin{array}{cccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right] \end{gathered}$$

Hence, the orthogonal projection of the vector$\overrightarrow{x}$ is$pro{j}_v\overrightarrow{x}=\left[\begin{array}{cccc}\frac{3}{4}& \frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right]$ .